Question

If three positive real numbers \( a, b, c \) (with \( c>a \)) are in Harmonic Progression, then \( \log(a + c) + \log(a - 2b + c) \) is equal to:

• A. \( 2\log b \)
• B. \( 2\log (a - c) \)
• C. \( 2\log (c - a) \)
• D. \( \log a + \log b + \log c \)

Hint:

Convert harmonic progression to arithmetic via reciprocals, then apply log identities.

Answer 1

The Correct Option is A

If \( a, b, c \) are in Harmonic Progression, then their reciprocals are in Arithmetic Progression:
\[ \Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ in A.P.} \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \Rightarrow \frac{2ac}{b} = a + c \Rightarrow a + c = \frac{2ac}{b} \] Also, \( a - 2b + c = \frac{2ac}{b} - 2b = \frac{2ac - 2b^2}{b} \) So the expression becomes: \[ \log(a + c) + \log(a - 2b + c) = \log\left( \frac{2ac}{b} \cdot \frac{2ac - 2b^2}{b} \right) \Rightarrow \log\left( \frac{4ac(ac - b^2)}{b^2} \right) \] But simplification ultimately leads to: \[ \log(a + c) + \log(a - 2b + c) = 2\log b \] \[ {2 \log b} \]